Timeline for Closed paths, traces and spectra
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2020 at 11:32 | comment | added | Michael Magee | to bound roots of a polynomial in terms of its coefficients is in general extremely difficult. | |
| Jun 24, 2020 at 11:31 | comment | added | Michael Magee | The intuitive difference between this and Ihara zeta is that Ihara involves reciprocal of a characteristic polynomial and hence by Cauchy identity is related to $\mathrm{Sym}^k A$ whereas $\mathrm{Tr} \bigwedge^k A$ shows up in the characteristic polynomial. So closed paths are telling you something about the coefficients of the characteristic polynomial. But there is also a formula for log of characteristic polynomial (in some region of parameter) in terms of a power series involving all the $\mathrm{Tr} (A^k)$. I don't know if any of this is helpful... of course trying... | |
| Jun 24, 2020 at 11:25 | comment | added | Michael Magee | More precisely, $\mathrm{Tr}\bigwedge^k A$ has a term for each vertex disjoint collection of closed (oriented) paths (as above) whose lengths sum to $k$ (every term comes from a permutation in $\mathrm{Sym}_k$ and the cycle type of this permutation dictates the lengths of the paths). The term also contains the sign of the permutation it comes from: this probably makes it useless for analytic purposes unless you know there are no closed paths of certain lengths. | |
| Jun 23, 2020 at 21:37 | comment | added | H A Helfgott | can you actually write down $\mathrm{Tr} \bigwedge A^{k}$ in terms of the number of closed paths in the sense of closed paths given above (as opposed to the weaker sense used in the definition of an Ihara zeta function)? I don't doubt it can be done - I simply want to make sure there are no grounds for confusion. | |
| Jun 23, 2020 at 5:44 | comment | added | H A Helfgott | (I obviously meant $\mathrm{Tr} A^{2k}\geq \sum_i \alpha_i^{2 k}$ in the above.) | |
| Jun 23, 2020 at 5:38 | comment | added | H A Helfgott | Also, what exactly do you mean when you say that "closed paths of length ≤k show up when you take the trace of the alternating kth power of A"? It's not the case (as I naïvely understood at first) that the trace is simply the number of closed paths of length $k$ (times $-1$), though that's certainly a term. | |
| Jun 23, 2020 at 4:10 | comment | added | H A Helfgott | @Michael_Magee [cont'd] However, even if we work with $\bigwedge^k A^2$ instead of $\bigwedge^{2 k} A$, we do not necessarily have $\mathrm{Tr} \bigwedge^k A^2 \geq \sum_{i_1<i_2<\dotsc<i_k} \alpha_{i_1}\dotsc \alpha_{i_k}$ (do we?), so I'm in a bit of a pickle. | |
| Jun 23, 2020 at 4:02 | comment | added | H A Helfgott | @Michael_Magee What happens if, instead of the alternating $k$th power of $A$, we take the symmetric $k$th power of $A$? I am asking because, on the spectral side, I am working with real orthonormal vectors $v_i$ that satisfy $\alpha_i:=\langle v_i, A v_i\rangle$ large (at least for many $i$) without necessarily being eigenvectors. When I consider $\mathrm{Tr} A^{2k}$, all is fine, since $\mathrm{Tr} A^{2k} \geq \sum_i \alpha_i$ gives me a useful lower bound on $\mathrm{Tr} A^{2k}$. (continued in next comment) | |
| Jun 23, 2020 at 0:34 | comment | added | Michael Magee | I added some more information: after thinking about it I realized it is not exactly trails that show up but a weaker variant: I added the details anyway. | |
| Jun 23, 2020 at 0:28 | history | edited | Michael Magee | CC BY-SA 4.0 |
added explanation of comment on trails at OPs request
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| Jun 22, 2020 at 18:48 | comment | added | H A Helfgott | Thank you! Can you go into a bit more detail on your last comment - on trails? | |
| Jun 22, 2020 at 18:20 | review | First posts | |||
| Jun 22, 2020 at 18:32 | |||||
| Jun 22, 2020 at 18:12 | history | answered | Michael Magee | CC BY-SA 4.0 |